A382424 The number of exponents in the prime factorization of n-th biquadratefree number that are equal to 3.

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0

Offset

1

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Sourabhashis Das, Wentang Kuo, and Yu-Ru Liu, On the number of prime factors with a given multiplicity over h-free and h-full numbers, Journal of Number Theory, Vol. 267 (2025), pp. 176-201; arXiv preprint, arXiv:2409.11275 [math.NT], 2024. See Theorem 1.2.

Formula

a(n) = A295883(A046100(n)).

a(n) = A382425(n) - A382423(n).

Sum_{A046100(k) <= x} a(k) = c * x + O(x^(1/3)/log(x)), where c = (1/zeta(4)) * Sum_{p prime} ((p-1)/(p^4-1)) = 0.094478403261541991612... (Das et al., 2025).

Sum_{k=1..n} a(k) ~ c * n, where c = Sum_{p prime} ((p-1)/(p^4-1)) = 0.102256170933897073... - Vaclav Kotesovec, Mar 25 2025 (according to the above formula)

Mathematica

f[k_] := Module[{e = If[k == 1, {}, FactorInteger[k][[;; , 2]]]}, If[AllTrue[e, # < 4 &], Count[e, 3], Nothing]]; Array[f, 150]

Prog

(PARI) list(kmax) = {my(e, is); for(k = 1, kmax, e = factor(k)[, 2]; is = 1; for(i = 1, #e, if(e[i] > 3, is = 0; break)); if(is, print1(#select(x -> x == 3, e), ", "))); }

Crossrefs

Cf. A013662, A046100, A295883, A376366, A382422, A382423, A382425.

Sequence in context: A295662 A367512 A382290 * A187946 A330549 A380439

Adjacent sequences: A382421 A382422 A382423 * A382425 A382426 A382427

Keyword

nonn,easy

Author

Amiram Eldar, Mar 25 2025

Status

approved